We have seen the argument for proving the above statment using Zorn's Lemma by asserting the existence of a maximal linearly independent set which serves a basis. In finite dimensional vector space, a basis is same as a maximal linearly independent and also same as a minimal spanning set.

Does the notion of minimal spanning set make sense for arbitrary vector spaces?

Moreover, can the statement that every vector space has a basis be proved using the partially ordered set $\Sigma = \lbrace A \subset V \vert Span(A) =V \rbrace $ with the partial order $A \leq B$ iff $B \subset A $?

Can one say intersection of a chain of spanning sets in this poset is also a spanning set?

Does the notion of minimal spanning set make sense for arbitrary vector spaces?

Sure: why not? The notion has a sensible definition. A minimal spanning set $S$ is one for which $\langle S'\rangle \subsetneq \langle S\rangle $ whenever $S'$ is a proper subset of $S$.

If instead $S'$ were a proper subset of $S$, and yet $\langle S'\rangle = \langle S\rangle $, then it would follow that every element of $S\setminus S'$ is a linear combination of the elements of $S'$, and hence linearly dependent upon the elements of $S'$. So, elements could be removed from $S$ while preserving the span, and $S$ would not be a minimal spanning set.

..can the statement that every vector space has a basis be proved using the partially ordered set $\Sigma = \lbrace A \subset V \vert Span(A) =V \rbrace $ and define partial order $A \leq B$ iff $B \subset A $. Can one say intersection of all totally ordered spanning set is also a spanning set?

(Prism's answer tells why the answer is negative.)

So in summary, the poset of spanning sets ordered by inclusion (and the poset of spanning sets ordered by reverse-inclusion) have perfectly well-defined minimal (respectively maximal) elements. It's just that the hypotheses of Zorn's lemma are not satisfied, and therefore it can't successfully be applied to deduce the existence of such minimal (resp. maximal) elements.